\newproblem{lay:7_4_19}{
  % Problem identification
	\begin{large}
	  \hspace{\fill}\newline
    \textbf{Lay, 7.4.19}
	\end{large}
	\\
  \ifthenelse{\boolean{identifyAuthor}}{\textit{Carlos Oscar Sorzano, Aug. 31st, 2013} \\}{}

  % Problem statement
	Given a SVD decomposition of a matrix $A$, $A=U\Sigma V^T$, show that the columns of $V$ are eigenvectors of $A^TA$, the columns of $U$ are eigenvectors of $AA^T$,
	and the diagonal entries of $\Sigma$ are the singular values of $A$. [\textit{Hint}: Use the SVD to compute $A^TA$ and $AA^T$.]
}{
   % Solution
	Let us calculate $A^TA$
	\begin{center}
		$A^TA=(U\Sigma V^T)^T(U\Sigma V^T)=V\Sigma^TU^TU\Sigma V^T=V(\Sigma^T\Sigma)V^T)$
	\end{center}
	But this is an eigendecomposition of $A^TA$ because $V$ is an orthogonal matrix $V^T=V^{-1}$ and $\Sigma^T\Sigma$ is a diagonal $n\times n$ matrix with $r$ values $\sigma_i^2$ (being
	$\sigma_i$ the singular values of $A$ and $r$ the number of non-zero singular values of $A$) and $n-r$ zeros. By the Diagonalization Theorem (Theorem 5.3.5), we have that the
	columns of $V$ are the eigenvectors of $A$ and the diagonal entries of $\Sigma^T\Sigma$ their corresponding eigenvalues.
	
	Since the eigenvalues of $A^TA$ are $\sigma_i^2$, $\sigma_i$ are the singular values of the matrix $A$.
	
	We can proceed analogously with $AA^T$
	\begin{center}
		$AA^T=(U\Sigma V^T)(U\Sigma V^T)^T=U\Sigma V^TV\Sigma^T U^T=U(\Sigma\Sigma^T)U^T)$
	\end{center}
	Similarly, the columns of $U$ are the eigenvectors of the matrix $AA^T$ and $\Sigma\Sigma^T$ is an $m\times m$ diagonal matrix with the eigenvalues of $AA^T$ ($r$ of them are non-zero
	and $m-r$ are zero).
}
\useproblem{lay:7_4_19}
\ifthenelse{\boolean{eachProblemInOnePage}}{\newpage}{}

